Centripetal Force and Uniform Circular Motion
Introduction
This lab gives ideas of the uniform circular motion. This has a different concept
from the previous labs, which are based on linear kinematics. Unlike a linear
motion, the circular motion is always changing its direction although the speed is
constant. The change of direction in a circular motion generates acceleration that
is directed to the center of the motion. As we have already learned, acceleration
acquires a force, which is, in circular motion, called centripetal force. In nature,
we can find this all of the places, such as driving a car into curved roads, planets
revolving about sun, etc. The centripetal force can be quantified as
Fc =
mv 2
r
where m is the mass of an object, v is the linear (or tangential) velocity, and r is the radius of the
motion. From the expression, the force is proportional to the velocity squared and inversely
proportional to the radius. (That’s why you have to slow down before driving into a curve!) In a
uniform circular motion, time repeated for one cycle is constant. The time for one revolution is
called period and the expression is:
T=
2πr
v
In terms of kinematics, time is defined as distance
divided by velocity. The distance for a circle is the
circumference, which is 2πr . Therefore, the period
can be expressed as above. The unit of period is
second (s).
In the experiment, the radius of orbit (indicated by the
pointer) and the period is measured by a ruler and a
photo gate, respectively, and the linear velocity is
calculated as
v=
2πr
T
The centripetal force obtained by this experiment (See the figure above) is determined by the
spring. Spring force depends on the stretched length, which is associated with the radius, r. When
the spring is more stretched, it will attain more force, which will also be equal to the
corresponding centripetal force.
To calibrate the force, gravitational force is used when the system is static. The centripetal force
(or centrifugal force which is the re-action force of centripetal force) can be calibrated by the
hanging mass, Fg = Mg :
Fc = Fg
(This equation is only for the above setting.)
Objectives:
• To understand the property of the centripetal force
• To learn how to measure the centripetal force by imitating the centrifugal force (reaction force of the centripetal force) with the gravitational force.
• To find how mass of the object affects its centripetal force.
Check the notations and names of parts of the equipment.
Question before the experiment:
What are the period, tangential speed, and centripetal force? Please write them down in layman’s
terms, or try to rephrase what your teacher explains.
The procedure and data acquisition
⇐ Start up DataStudio. Click “Create Experiment.” Click on the digital
channel to select “Photogate.”
Click “Setup Timers.” ⇒
Click
this.
⇐ Click “Timing Sequence Choices” to select
“Blocked.”
Repeat the above procedure to have two of the “Ch 1
Blocked” in the display. Then, click “Done.” ⇒
⇐ Choose “Table” for “Timer 1 (s).”
Each “Elapsed Time” is the period of the uniform circular motion. ⇒
The experimental procedure [Check each variables with the figure in introduction. They are case
sensitive.]
1.
Set up the position of index point.
⇐ Measure the radius, r, and record it in the table. It
is from the center of the axis to the pointer.
Warning
Make sure if the bob and pointer is lined up when you
remove from the spring.
2. Spin the bar and measure the period.
Then, rotate the vertical bar. (Use the grip.) After the pointer and tip
of the bob are lined up (or when the bob is touching the pointer), make
sure that the recorded period is stable. [Record the data of period,
T.]⇒
3. Stop rotation and hang a weight.
⇐ To find the centripetal force, Fc, you have to
calibrate it with an equivalent force. The
centripetal force is equal to the centrifugal
one in this system, which can stretch the
spring to the top of the tip. To calibrate it,
you may use the gravitational force. Hang
masses to the other side until the bob and
pointer are lined up. The force will be the
mass × gravitational acceleration which is
supposed to be equal to the centripetal force.
Question:
Do you understand why you have to hang masses as above? Yes_____ No _______
If you choose no, ask your teacher to be convinced.
1. Measurement of Centripetal Force (for the different-mass object with a fixed
radius)
Case 1: m = only bob
r (Radius of orbit) = ____________ (m)
Mass of only bob = _________________ (kg)
After making stable circular motion, record about 6 periods. Remove
mistakenly obtained values. Go to “Table” and click the icon, Σ.
Write the mean value of period and calculate the difference between
maximum and minimum data |Max. – Min.| as the fluctuation of data.
Click this icon to obtain the
mean, max. and min. values.
m
T
Total mass of
object
Average
(mean) period
1
Only mass of bob
2
Only mass of bob
3
Only mass of bob
|Max. – Min.|
Fluctuation of
data of period
v = 2πr T
Fc = mv 2 r
Tangential
speed
Centripetal force
M
Hanging mass
(This is not the
mass of bob!)
Fg = Mg
(g = 9.807 m/s2)
calibrated force
% difference
b/w Fc and
Fg
The mass hanger itself is 0.050 kg to add. 50 g
Average and standard deviation of ( Fc ):
±
(
)(
) ⇐ write the unit
Case 2: m = bob + 200 g
r (Radius of orbit) = ____________ (m)
m
T
Total mass of
object
(bob + added)
Average
(mean) period
1
Add 0.200 kg
2
Add 0.200 kg
3
Add 0.200 kg
|Max. – Min.|
Fluctuation of
data of period
Mass of only bob = _________________ (kg)
v = 2πr T
tangential speed
Fc = mv 2 r
centripetal force
M
hanging mass
(This is not the
mass of bob!)
Fg = Mg
(g = 9.807 m/s )
calibrated force
Average and standard deviation of ( Fc ):
(
±
)(
2
) ⇐ write the unit
% difference
b/w Fc and
Fg
2. Measurement of Centripetal Force (for the different-radius with a fixed mass)
[Follow the same procedure as previous except calculating standard deviation.]
Case 1: The r is shorter. (Do not copy the result from the first part. The average period
and its fluctuation may be different.)
r (Radius of orbit) = ____________ (m)
1
m
T
Total mass of
object
Average
(mean) period
|Max. – Min.|
Fluctuation of
data of period
Mass of only bob = _________________ (kg)
v = 2πr T
Tangential
speed
Fc = mv 2 r
Centripetal force
M
Hanging mass
(This is not the
mass of bob!)
Fg = Mg
2
(g = 9.807 m/s )
calibrated force
% difference
b/w Fc and
Fg
Only mass of bob
Case 2: The r is longer. [Warning: This is important to obtain accurate results. Follow the
instruction when you change the radius.]
Take the spring off from the bob.
For the new radius, adjust the
bar so the bob can be lined up
with the new location of
index pointer. Hook the spring to
the bob again and do the
same procedure as previous.
r (Radius of orbit) = ____________ (m)
1
m
T
Total mass of
object
Average
(mean) period
Only mass of bob
|Max. – Min.|
Fluctuation of
data of period
Mass of only bob = _________________ (kg)
v = 2πr T
Tangential
speed
Fc = mv 2 r
Centripetal force
M
Hanging mass
(This is not the
mass of bob!)
Fg = Mg
2
(g = 9.807 m/s )
calibrated force
% difference
b/w Fc and
Fg
3. Analysis and discussions
Question 1: In the first part, how does the centripetal force depend on the mass? In this set up,
mass is not supposed to affect the force. Write down the mass and its centripetal force and
compare them. Is there a large difference in the forces you obtained? (Is it large enough even
considering the standard deviation of the force?) Think about why. [Hint: How does the period
change according to the change of mass?]
Question 2: In the second part, how does the centripetal force depend on the radius? Discuss this
referring to the underlined sentences in the introduction.
Questions 3: For the following cases, the formula, Fc=mv2/r, illustrates the relationships between m, v,
and r: When a trailer and a car go into each curve whose radius is 15.0 m, the trailer will obtain more
centripetal force because of more mass than the car. Similarly, the shorter radius the curve has the more
force the car attains as far as the other conditions are equal. Similarly, the more velocity the car takes the
more force the car attains with equal mass and radius.
For this lab, strength of the centripetal force is fixed by the distance stretched by the spring; namely, it is
constant. Because of that, the answer of Question 1 is not straightforward as you simply look at Fc=mv2/r.
List other daily life circular motions, such as planetary motions around sun, etc. How are the centripetal
force, mass, radius and period of the motion related each other? Which parameters are constant? Can you
simply use the formula, Fc=mv2/r?
Questions you want to explore: Address this to your report!
What are the possible errors in this lab? How do you improve the experimentation to obtain more
accurate results? Why is your idea more effective? [This part must come from your observation and
insights with the experimental results. Write also possible reasons. Discussion with your partners or TA is
highly encouraged.]